I found many implementations that deal with splitting a polygon by a given line, but I only need to split a Quad (rectangle with 4 vertexes).
Is there an algorithm optimized for this task? Simplicity is valued over performance.
I narrowed down 4 types of intersection:
Adjacent
Where the line enters one side and leaves through an adjacent side.
This will generate 1 polygon with 3 points and 1 polygon with 5 points.
Opposite
Where the line enters one side and leaves through the opposite side.
This will generate 1 polygon with 4 points and 1 polygon with 4 points.
Diagonal Opposite
Where the line enters one corner and leaves through the opposite corner.
This will generate 1 polygon with 3 points and 1 polygon with 3 points.
Diagonal Adjacent
Where the line enters one corner and leaves through an adjacent side.
This will generate 1 polygon with 3 points and 1 polygon with 4 points.
But so far I was unable to come up with a good simple algorithm.
There are a lot of effective algorithms to clip a line by rectangular window.
I've used Liang-Barski one for my purposes (check "External Links" section for effective implementation)
Related
How to place n points in a plane so that the distance between every two points is unique and at the same time we could choose one of them that is the closest for all others n-1 points.
I tried to draw it. But I was able to draw it only for max n=5. I drew two perpendicular lines. One point I was intersection of the lines and the others points lie on the line so that they formed a quadrilateral and point I was inside. With more points it seems impossible to me but I can't prove why
As the pic show, both curves are of spline line, and have limited points. I want to figure out the count of green points. Is there any idea?
I assume the black curve is x-monotone (otherwise the "one above the other" term can be ambiguous).
A simple approach is to consider the black curve as a polygonal line and for each point p on the red curve find the point q on the polygonal line with the same x-coordinate. Then the green points are those p that have a larger y-coordinate than their corresponding q.
Finding the point q corresponding to a given p amounts to going over the segments of the polygonal line and identifying segments that have one endpoint with smaller x-coordinate and the other with larger. Once you have such a segment the y-value of q is just a linear interpolation.
Since the polygonal line is x-monotone, the x-coordinates of the points are sorted. Therefore, the search for the corresponding segments can be done efficiently using logarithmic binary-search.
In my current project I need to calculate the intersection area of triangles and the unit squares in an infinite grid.
For every triangle (given by three pairs of floating point numbers) I need to know the area (in the interval (0,1]) it has in common with every square it intersects.
Right now I convert both (the triangle and the square) to polygons and use Sutherland-Hodgman polygon clipping to calculate the intersection polygon, which I then use to calculate its area.
This approach now shows to be a performance bottleneck in my application. I guess a more specialized (analytical) algorithm would be much faster. Is there a standard solution for this problem, or do you have any idea? I only need the areas, not the shape of the intersections.
Your polygon are convex. There are some algorithms for convex polygons faster than general ones. I've used O'Rourke algorithm with success (code from his book here, I believe that good description exists). Note that some values may be precomputed for your squares.
If your polygons not always intersect, then you may at first check the fact of intersection with separating axes method.
Another option to try- Liang-Barski algorithm for clipping every triangle edge by square.
Edit: You can quickly find all intersections of triangle edges with grid using algorthm of Amanatides and Woo (example in grid traversal section here)
To process this task with hi performance , i suggest some modifications of
Vatti line sweep clipping.
http://en.wikipedia.org/wiki/Vatti_clipping_algorithm
Stepping from minimal Y vertex of your Triangle make such steps:
sort vertexes by Y coordinate
step Y higher to MIN(nextVertex.Y, nextGridBottom)
Calculate points of intersection of grid with edges.
Collect current trapezoid
repeat from step2 until vertex with highest Y coordinate.
Split trapezoids by X coordinate if required.
here is example of Trapezoidalization in X direction
http://www.personal.kent.edu/~rmuhamma/Compgeometry/MyCG/PolyPart/polyPartition.htm
It illustrate main idea of line sweep algorithm. Good luck.
You are not mentioning what precision you are looking for. In case you are looking for a analytical method, disregard this answer, but if you just want to do antialiasing I suggest a scanline edge-flag algorithm by Kiia Kallio. I have used it a few times and it is quite fast and can be set up for very high precision. I have a java implementation if you are interested.
You can take advantage of the regular pattern of squares.
I'm assuming the reason this is a bottleneck is because you have to wait while your algorithm finds all squares intersecting any of the triangles and computes all the areas of intersection. So we'll compute all the areas, but in batches for each triangle in order to get the most information from the fewest calculations.
First, as explained by others, for each edge of the triangle, you can find the sequence of squares that edge passes through, as well as the points at which it crosses each vertical or horizontal edge of a square.
Do this for all three sides, keeping a list of all the squares you encounter, but keep only one copy of each square. It may be useful to store the squares in multiple lists, so that all squares on a given row are all kept in the same list.
When you've found all squares the triangle's edges pass through, if two of those squares were on the same row, any squares between those two that are not in the list are completely inside the triangle, so 100% of each of those squares is covered.
For the other squares, the calculation of area can depend on how many vertices of the triangle are in the square (0, 1, 2, or 3) and where the edges of the triangle intersect the sides of the square. You can summarize all the cases in a few pencil-and-paper drawings, and come up with calculations for each one. For example, when an edge of triangle crosses two sides of the square, with one corner of the square on the "outside" side of the edge, that corner is one angle of a small triangle "cut off" by that edge of the larger triangle; use the points of intersection on the square's sides to compute the area of the small triangle and deduct it from the area of the square. If two points instead of one are "outside", you have a trapezoid whose two base lengths are found from the points of intersection, and whose height is the width of the square; deduct its area from the square. If three points are outside, deduct the entire area of the square and then add the area of the small triangle.
One vertex of the large triangle inside the square, three corners of the square outside that angle: draw a line from the remaining corner to the triangle's vertex, so you have two small triangles, deduct the entire square and add those triangles' areas. Two corners of the square outside the angle, draw lines to the vertex to get three small triangles, etc.
I'm phrasing this so that you always assume you start with the entire area of the square and reduce the area by some amount depending on how the edge of the triangle intersects the square. That way, in the case where the edges of the triangle intersect the square more than twice--such as one edge cuts across one corner of the square and another edge cuts across a different corner, you can just deduct the area cut off by the first edge, then deduct the area cut off by the second edge.
This will be a considerable number of special cases, though you can take advantage of symmetry; for example, you don't have to write the complete calculation for "cut off a triangle in one corner" four times.
You'll write a lot more code than if you just took someone's convex-polygon library off the shelf, and you will want to test the living daylights out of it to make sure you didn't forget to code any cases, but once you get it working, it shouldn't take much more effort to make it reasonably fast.
I have a set of 2D points, unorganized, and I want to find the "contour" of this set (not the convex hull). I can't use alpha shapes because I have a speed objective (less than 10ms on an average computer).
My first approach was to compute a grid and find the outline squares (squares which have an empty square as a neighbor). So I think I downsized efficiently my numbers of points (from 22000 to 3000 roughly). But I still need to refine this new set.
My question is : how do I find the real outlines points among my green points ?
After a weekend full of reflexions, I may have found a convenient solution.
So we need a grid, we need to fill it with our points, no difficulty here.
We have to decide which squares are considered as "Contour". Our criteria is : at least one empty neighbor and at least 3 non empty neighbors.
We lack connectivity information. So we choose a "Contour" square which as 2 "Contour" neighbors or less. We then pick one of the neighbor. From that, we can start the expansion. We just circle around the current square to find the next "Contour" square, knowing the previous "Contour" squares. Our contour criteria prevent us from a dead end.
We now have vectors of connected squares, and normally if our shape doesn't have a hole, only one vector of connected squares !
Now for each square, we need to find the best point for the contour. We select the one which is farther from the barycenter of our plane. It works for most of the shapes. Another technique is to compute the barycenter of the empty neighbors of the selected square and choose the nearest point.
The red points are the contour of the green one. The technique used is the plane barycenter one.
For a set of 28000 points, this techniques take 8 ms. CGAL's Alpha shapes would take an average 125 ms for 28000 points.
PS : I hope I made myself clear, English is not my mothertongue :s
You really should use the alpha shapes. Maybe use only green points as inputs of the alpha alpha algorithm.
I'm experimenting with a vector based graphics style with objects represented as series of line segments with a given width(it would probably be easier to think of these as rectangles). The problem is that these segments are connected at the center and leave a gap (shown below). I've determined that the most efficient way to cover this gap is simply to cover it with a triangle, and since I'm working in OpenGL, all I need are the points of the two points that don't overlap with the other rectangle, the third point being the center point where the two line segments(rectangles) are connected. How can I determine which points I need to use for the triangle, given that I have all of the points from both rectangles?
EDIT: I will also accept alternative solutions, as long as they cover up that gap.
EDIT 2: Nevermind, I solved it. I'll post code once I have better Internet connection.
Maybe I'm misunderstanding the question... but if you zoom in on the top corner of your red pentagon, you get something like this, am I right?
where A and B are nodes on the rectangle for edge1 and C and D are nodes on the rectangle for edge2. You say you already know these coordinates. And from what you say, the edges meet at the centre, which is halfway between A and B, and also halfway between C and D. So call this point X, and you can calculate its coordinates easily I guess.
So all you need to do is draw the missing triangle AXC, right? So one way would be to determine that A and C are on the "outside" of the polygon (and therefore need filling) and B and D are on the "inside" and therefore don't. But it's probably easier to just draw both, as it doesn't hurt. So if you fill AXC and BXD, you'd get this:
The solution I found assumes that there are 3 basic cases:
First, the three unique center points for the two rectangle proceed upward (positive y direction) so the gap is either on the left or right of the connection. In my code, I had the corner points of the rectangle organized by their orientation to the left or right of the center point, so if the bottom rectangle's left point is below the top rectangle's left point, then the gap is between the left points of the two rectangles, otherwise the gap is between the right points.
Second, the three unique center points have a maximum at the center most of the center points, so the gap is on the top. The gap is then between the two points with the maximum y values.
Third, the three unique center points have a minimum at the center most of the center points, so the gap is on the bottom. The gap is then between the two points with the minimum y values.
[I'll post pictures of the example cases if it is requested]